1 + frac{2}{3!} + frac{3}{5!} + frac{4}{7!} + | vedclass

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(C) The given series is $S = \sum_{n=1}^{\infty} \frac{n}{(2n-1)!}$.We can rewrite the general term $T_n$ as:$T_n = \frac{n}{(2n-1)!} = \frac{1}{2} \c

Vedclass · Accepted Answer

$e/2$

Vedclass · Answer

(C) The given series is $S = \sum_{n=1}^{\infty} \frac{n}{(2n-1)!}$.We can rewrite the general term $T_n$ as:$T_n = \frac{n}{(2n-1)!} = \frac{1}{2} \cdot \frac{2n}{(2n-1)!} = \frac{1}{2} \cdot \frac{(2n-1) + 1}{(2n-1)!}$$T_n = \frac{1}{2} \left( \frac{2n-1}{(2n-1)!} + \frac{1}{(2n-1)!} \right) = \frac{1}{2} \left( \frac{1}{(2n-2)!} + \frac{1}{(2n-1)!} \right)$.Summing from $n=1$ to $\infty$:$S = \frac{1}{2} \left( \sum_{n=1}^{\infty} \frac{1}{(2n-2)!} + \sum_{n=1}^{\infty} \frac{1}{(2n-1)!} \right)$.Recall the series expansions:$e = \sum_{k=0}^{\infty} \frac{1}{k!} = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots$$e^{-1} = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} = \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots$Thus,$\sum_{n=1}^{\infty} \frac{1}{(2n-2)!} = \frac{1}{0!} + \frac{1}{2!} + \frac{1}{4!} + \dots = \frac{e + e^{-1}}{2}$.And $\sum_{n=1}^{\infty} \frac{1}{(2n-1)!} = \frac{1}{1!} + \frac{1}{3!} + \frac{1}{5!} + \dots = \frac{e - e^{-1}}{2}$.Substituting these values:$S = \frac{1}{2} \left( \frac{e + e^{-1}}{2} + \frac{e - e^{-1}}{2} \right) = \frac{1}{2} \left( \frac{2e}{2} \right) = \frac{e}{2}$.

$1 + \frac{2}{3!} + \frac{3}{5!} + \frac{4}{7!} + \dots \infty = \,$

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